Let a circle of radius 4 be concentric to the ellipse $15 x^{2}+19 y^{2}=285$. Then the common tangents are inclined to the minor axis of the ellipse at the angle :

We have, equation of ellipse : $15 x^2+19 y^2=285$ <br/><br/>or $ \frac{x^2}{19}+\frac{y^2}{15}=1$ <br/><br/>Let the coordinate of center of circle be $(0,0)$. <br/><br/>Equation of circle is $x^2+y^2=16$ <br/><br/>Equation of tangent of ellipse is <br/><br/>$$ \begin{gathered} y=m x \pm \sqrt{19 m^2+15} \text { or } \\\\ m x-y \pm \sqrt{19 m^2+15}=0 \end{gathered} $$ <br/><br/>It is also tangent to the circle $x^2+y^2=16$ <br/><br/>Perpendicular distance from center of circle to tangent $=4$ <br/><br/>$\frac{\left|0-0 \pm \sqrt{19 m^2+15}\right|}{\sqrt{m^2+1}}=4$ <br/><br/>On squaring both side, we get <br/><br/>$$ \begin{gathered} 19 m^2+15=16 m^2+16 \\\\ 3 m^2=1 \\\\ m^2=\frac{1}{3} \\\\ m= \pm \frac{1}{\sqrt{3}} \end{gathered} $$ <br/><br/>$$ \tan \theta=\frac{1}{\sqrt{3}} \Rightarrow \theta=\frac{\pi}{6} \text { with } X \text {-axis or } $$$\frac{\pi}{3}$ with $Y$-axis <br/><br/>Hence, the common tangents are inclined to the minor axis of the ellipse at an angle of $\frac{\pi}{3}$.